Answer
$$2{e^{\sqrt {x - 1} }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{\sqrt {x - 1} }}}}{{\sqrt {x - 1} }}} dx \cr
& {\text{substitute }}u = \sqrt {x - 1} \cr
& du = \frac{1}{{2\sqrt {x - 1} }}dx \cr
& 2du = \frac{1}{{\sqrt {x - 1} }}dx \cr
& = \int {{e^{\sqrt {x - 1} }}\frac{1}{{\sqrt {x - 1} }}} dx \cr
& \int {{e^u}\left( 2 \right)} du \cr
& = 2\int {{e^u}du} \cr
& {\text{find the antiderivative }} \cr
& = 2{e^u} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \sqrt {x - 1} \cr
& = 2{e^{\sqrt {x - 1} }} + C \cr} $$
